This Exam P sample reference tests Exponential Distribution. Each policy has mean and standard deviation 1000. For 100 policies, premiums exceed the claim mean by one aggregate standard deviation, so the loss probability is 0.1587, choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.001 is a far upper-tail probability and is incompatible with the computed cutoff z=1.
Original practice · fully worked
Original variant: aggregate warranty-cost overrun
Warranty cost per device has mean 80 dollars and standard deviation 50 dollars. A seller ships 225 independent devices and collects a reserve of 18,750 dollars. Approximate the probability total warranty cost exceeds the reserve.
C 0.0228
B 0.0668
A 0.1587
D 0.3085
E 0.5000
Variant answer in brief
The total mean is 18,000 and standard deviation 750. The reserve is one standard deviation above the mean, leaving upper-tail probability 0.1587.
Setup
Setup
Let S be the total warranty cost for 225 independent devices, each with mean 80 dollars and standard deviation 50 dollars.
S=i=1∑225Xi
Model
Model
The aggregate mean is 18,000 dollars and the aggregate standard deviation is 750 dollars.
E[S]=225(80)=18000
SD(S)=50225=750
Compute
Compute
The reserve 18,750 is one standard deviation above the mean, so the approximate overrun probability is the standard normal upper tail at 1.
z=75018750−18000=1
P(S>18750)≈1−Φ(1)=0.1586553
Answer
Answer
The overrun probability is approximately 0.1587, corresponding to choice A.
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